math-optimal diameterand height
posted by Alicia on .
A cylindrical oil-storage tank is to be constructed for which the following costs apply:
cost per square meter metal for ides $30.00
combined costs of concrete base and metal bottom $37.50(cost per square meter)
top 7.50 (cost per square meter)
The tank is to be constructed with dimensions such that the cost is minimum for whatever capacity is selected.
a) One possible approach to selecting the capacity is to build the tank large enough for an additional cubic meter of capacity to cost $8.(note that this does not mean $8 per cubic meter average for the entire tank.) what is the optimal diameter and optimal height of the tank?
The last part of the question confuses me, I will solve the question in the traditional way
Let the radius be r m, the height be h m and the volume be V m^3, where V is a constant
We know V = πr^2h
h = V/(πr^2)
Cost = 7.5(top area) + 37.5(bottom area) + 30(side area)
= 7.5πr^2 + 37.5πr^2 + 30(2πr)(h)
= 45πr^2 + 60V/r
d(Cost)/dr = 90πr - 60V/r^2 = 0 for a min of Cost
90πr = 60V/r^2
r^3 = 2(πr^2h)/(3π)
3r = 2h
The ration of radius to height has to be 2 : 3
I will let you deal with the weird condition at the end.
After doing a "search" for your question, I noticed that
MathMate had answered the same question in a much more detailed way.